Industrial Engineering - Operations Research

The field of operations research provides a more powerful approach to decision making than ordinary software and data analytics tools. Employing operations research professionals can help companies achieve more complete datasets, consider all available options, predict all possible outcomes and estimate risk.

Additionally, operations research can be tailored to specific business processes or use cases to determine which techniques are most appropriate to solve the problem.

In this course we are going to get the knowledge of various tools & practices of operations research with an ultimate aim to optimize problems & increase productivity. The course is divided into several sections, Linear Programming Problems, Transportation problems, Assignment problems, Sequencing, queuing theory & lastly PERT & CPM.

Problems which seek to maximize (or, minimize) profit (or, cost) form a general class of problems called optimization problems. Thus, an optimization problem may involve finding maximum profit, minimum cost, or minimum use of resources etc.

A special but a very important class of optimization problems is linear programming problem.

Linear Programming Problem is an special case of optimization problems. Let's understand its mathematical formulation with the help of a very simple example.

Steps to solve an LPP:

1. Formulate the problem or LPP Formulation

2. Draw the graph including all constraints.

3. Find feasible region & check corner points.

This lecture is tailored to make you understand how LPP is formulated with the help of appropriate table, constraints & objective function.

I know that many of you must be knowing how to draw graphs. Its elementary knowledge but still some students forgets. Let’s take an very simple example to learn how to draw graphs. Just to brush up your memory.

Numerical - A company produces two types of toys: P & Q. Production time of Q is twice that of P & the company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (Of any type) per day. Toy type Q requires an electric switch which is available at 600 pcs/day only. The company makes a profit of 3 & 5 denominations for P & Q respectively.

What should be the daily production quantity of P & Q for maximization of profit?

ANS - 1000 & 500 units of P & Q respectively.

There is a shortcut method for finding optimum solution without having the need to make the graph. We just need to find out the slope of all the equations involved.

Binding constraint is an equation in linear programming that satisfies the optimal solution through its value. Finding the satisfactory optimal solution through the certain value by using the equation in linear programming is known as a binding constraint.

In simple language, A binding constraint is one whose value will satisfy all the constraints.

Special Cases

1. Unique Solution

2. Multiple Optimum Solution (More than one solution)

3. No solution

4. Unbounded Solution

Graphical method is an easier way to solve linear programming problems & even easier method we have just discussed was slope method. But these 2 methods are restricted to only 2 variables.

So here comes the next optimization method – Simplex. The simplex method can be utilized to optimize problems having 2 or more than 2 variables. Simplex method is a step by step procedure in which we improve solution in every step. This step by step improvement of solution is known as iterative process. It’s a bit lengthier than graphical method but solution is more optimized.

It follows the same initial procedure as in case of graphical method of finding out objective function & writing constraints but instead of making graphs, we make tables.

The steps involved in simplex methods are – Eradicating inequalities, Finding the basic feasible solution, making tables & iteration till most optimized solution is reached. We are going to understand the first two steps in this lecture & the last two with the help of a numerical in the next lecture.

Numerical - A toy manufacturing unit produces four-wheel cars & three-wheel auto. The raw material used for making these two toys are Plastic body & wheels. 2 Kgs of plastic material is utilized for toy car & 1 Kg is utilized for toy auto and 4 wheels are utilized for toy car & 3 wheels are utilized for toy auto. The toy car gives profit per unit as 7 denominations & toy autos is giving profit per unit as 5 denominations. Find out the optimum number of toy car & auto to be manufactured to get maximum PROFIT.

Ans - The maximum profit will come from selling 30 toy cars & 40 toy autos.

Transportation problem is a special kind of Linear Programming Problem (LPP) in which goods are transported from a set of sources to a set of destinations subjected to the supply and demand of the sources and destination such that the total cost of transportation is minimized.

A typical transportation table (As discussed in the lecture), is attached herewith for your reference.

Transportation problem works in a way of minimizing the cost function.The cost function is the amount of money spent to the logistics provider for transporting the commodities from production or supplier place to the demand place. It includes the distance between the two locations, the path followed, mode of transport, the number of units that are transported, the speed of transport, etc.

Allocations are the most appropriate placement of resources such that we get an optimized way to carry on transportation with minimum loss to cost function..

Initial basic feasible solution through N-W corner rule can be obtained by following certain steps. These steps are comprehensively explained in the pdf attached with this lecture. Please give it a look before attending.

VAM is a preferred method over earlier methods due to its solution being either optimal or very near to optimal. This method yields an initial basic feasible solution which may reduce the time for optimal calculations. The various steps involved in this method is explained in the attached PDF. Kindly go through it before attending this lecture.

Numerical - A product made in two factories P & Q, is transported to two destination R & S.Factory P produces 7 units & factory Q produces 9 units of the product. Each destination requires 8 units. If the NW corner method provides the total transportation cost as “X” & the optimized solution from VAM method gives transportation cost as “Y”. Then find the value of X – Y.

Ans - 27 Denomination

Assignment problem is a special type of linear programming problem which deals with the allocation of the various resources to the various activities on one to one basis. It does it in such a way that the cost or time involved in the process is minimum and profit or sale is maximum. Though these problems can be solved by simplex method or by transportation method but assignment model gives a simpler approach for these problems.

In assignment problems, the matrix is always a square one that means m = n & only one job is assigned to every assignee. In simpler terms, one destination, one supply. Unlike transportation problems where one destination can have several suppliers.

The Hungarian Method is based on the principle that if a constant is added to every element of a row and/or a column of cost matrix, the optimum solution of the resulting assignment problem is the same as the original problem and vice versa.

Numerical - A firm is required to procure 3 items P, Q & R. The price quoted for these items by suppliers are S1, S2 & S3 respectively. The prices are given in the table (In lecture). The management policy requires that each item has to be supplied by only one supplier and one supplier supply only one item. What would be the minimum total cost of procurement to the firm.

Ans - P = 120, Q = 140 &, R = 125 denominations

Whenever you find a maximization problem that means if you need to maximize something in terms of a transportation or assignment problem, you need to convert the problem into a minimization one. To do that, just subtract each cell value from maximum cell value and write the corresponding values in each cell.

Sequencing is the selection of an appropriate order in which a number of jobs (Operations) can be assigned to a finite number of service facilities (Machines or equipment) so as to optimize the outputs in terms of time, cost or profit.

In case of sequencing, we are solving for n no of jobs for n no of machines. Its completely different than the transportation & assignment problems.

The ultimate aim of sequencing is idleness reduction. This method increases productivity, profit & efficiency of a machine exponentially if applied correctly.

Shortest processing time (SPT) rule would require the job with the shortest processing time to be processed ahead of all other waiting jobs. The use of this rule for sequencing will minimize the average time a job spends at the workcenter (waiting time plus processing time), which is called average time in the system.

EDD (Earliest Due Date). EDD sequencing rule is that “jobs are processed according to the due date, earliest due date first.”

To find the order for n jobs in 2 machines, we apply Johnson’s algorithm. Its primary objective is to find an optimal sequence of jobs to reduce make span (the total amount of time it takes to complete all jobs). It also reduces the amount of idle time between the two work centres.

Numerical - A manufacturing shop processes sheet metal jobs where each job must pass through 2 machines M1 & M2. In that order. The processing time (in hrs) for these jobs are given in the table (In lecture). What is the optimal MST (Make Span Time) in hours of the shop? Also find the % utilization of each of the machinery.

Ans - MST = 115 Hours

% utilization of M/C 1 = Total M/C time – Idle time / MST = 109/115 = 94.7%

% utilization of M/C 2 = 115 – (8+11+12+5) / 115 = 68.6%

Queuing theory refers to the mathematical study of the formation, function, and congestion of waiting lines, or queues.

At its core, a queuing situation involves two parts.

1. Someone or something that requests a service — usually referred to as the customer, job, or request.

2. Someone or something that completes or delivers the services — usually referred to as the server.

The example used to elaborate the same (The Doctor's clinic one) is attached herewith.

Waiting in line is a part of everyday life because as a process it has several important functions. Queues are a fair and essential way of dealing with the flow of customers when there are limited resources. Negative outcomes arise if a queue process isn’t established to deal with overcapacity.

Queuing theory is important because it helps describe features of the queue, like average wait time, and provides the tools for optimizing queues. From a business sense, queuing theory informs the construction of efficient and cost-effective workflow systems.

Queuing theory is powerful because the ubiquity of queue situations means there are countless and diverse applications of queuing theory.

Queuing theory uses the Kendall notation to classify the different types of queuing systems, or nodes. Queuing nodes are classified using the notation (a/b/c) : (d/e/f) where:

a = Arrival process/arrival rate

b = Mathematical distribution of the service time/Service rate

c = Number of servers

d = Service rule/Queuing discipline, assumed first-in-first-out (FIFO) if omitted

e = Capacity of the queue/System length, Omitted if unlimited

f = Number of possible customers/Calling population length, Omitted if unlimited

Most important formulas for solving queuing related problems are explained in this lecture.

Numerical - Customers arrives at a ticket counter at a rate of 50/hr and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1 minute. Assuming that customer arrival forms a Poisson process & service times are exponentially distributed, the average waiting time in queue in minutes is?

Ans - 5 Minutes